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                <h1 id="happens-before"><a href="#happens-before" class="headerlink" title="happens-before"></a>happens-before</h1><p>happens-before是JMM最核心的概念。并不是所有的重排序都要求有happens-before，jMM把重排序分为两种：1.会改变程序执行结果的重排序 2.不会改变程序执行结果的重排序。</p>
<p>对于会改变程序执行结果的重排序，JMM要求编译器和处理器必须禁止这种重排序。</p>
<p>对于不会改变程序执行结果的重排序，JMM对编译器和处理器不做要求。</p>
<h2 id="happens-before的定义"><a href="#happens-before的定义" class="headerlink" title="happens-before的定义"></a>happens-before的定义</h2><p><strong>JMM对程序员的承诺（可能不会）</strong>：如果一个操作happens-before另一个操作，那么第一个操作的执行结果将会对第二个操作可见，而且第一个操作的执行顺序排在第二个操作之前。</p>
<p><strong>JMM对编译器和处理器重排序的约束原则</strong>：两个操作之间存在happens-before关系，并不意味着java平台的具体实现必须要按照happens-before关系指定的顺序来执行。如果重排序之后的执行结果，与按happens-before关系来执行的结果一致，那么这种重排序并不非法（<strong>JMM允许这种不改变执行结果的重排序</strong>）。</p>
<h2 id="happens-before规则总结"><a href="#happens-before规则总结" class="headerlink" title="happens-before规则总结"></a>happens-before规则总结</h2><p>1.程序顺序规则：一个线程中的每一个操作，happens-before于该线程中的任意后续操作。</p>
<p>2.监视器锁规则：对一个锁的解锁，happens-before于随后这个锁的加锁。</p>
<p>3.volatile变量规则：对一个volatile域的写，happens-before于任意后续对这个volatile域的读。</p>
<p>4.传递性：如果A happens-before B, B happens-before C，A  happens-before C。</p>
<p>5.start()规则：如果线程A执行操作 ThreadB.start()操作,那么A线程的ThreadB.start()操作happens-before于线程B中的任意操作。</p>
<p>6.join()规则：如果线程A执行操作ThreadB.join()并成功返回，那么线程B中的任意操作happens-before于线程A从ThreadB.join操作成功返回。</p>
<p><strong>as-if-serial和happens-before的区别</strong></p>
<p>不同：as-if-serial语义是保证单线程内程序的执行结果不被改变，happens-before关系保证同步的多线程程序的执行结果不被改变。</p>
<p>相同：都是为了在不改变程序执行结果的前提下，尽可能地提高程序执行的并行度。</p>
<h1 id="双重检查锁定和延迟初始化"><a href="#双重检查锁定和延迟初始化" class="headerlink" title="双重检查锁定和延迟初始化"></a>双重检查锁定和延迟初始化</h1><p>在java多线程程序中，有时候需要采用延迟初始化来降低初始化类和创建对象的开销。双重检查锁定是常见的延迟初始化技术（错误，后续使用volatile进行修改）。</p>
<h2 id="双重检查锁定修改历史"><a href="#双重检查锁定修改历史" class="headerlink" title="双重检查锁定修改历史"></a><strong>双重检查锁定修改历史</strong></h2><pre class="line-numbers language-java" data-language="java"><code class="language-java">public class Singleton &#123;
    private static Singleton uniqueSingleton;

    private Singleton() &#123;
    &#125;

    public Singleton getInstance() &#123;
        if (null &#x3D;&#x3D; uniqueSingleton) &#123;
            uniqueSingleton &#x3D; new Singleton();
        &#125;
        return uniqueSingleton;
    &#125;
&#125;<span aria-hidden="true" class="line-numbers-rows"><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span></span></code></pre>

<p>实现单例模式时，如果没考虑多线程的情况下可能会导致。两个线程同时调用getInstance()方法，对象uniqueSingleton被两个线程实例化，并且被不同对象持有，这样就不是单例模式了。</p>
<p><img src="https://gitee.com/modest2001/img/raw/master/yygh/%E5%8F%8C%E9%87%8D%E6%A3%80%E6%9F%A5%E9%94%81%E5%AE%9A1.png" alt="image-20220404151502382"></p>
<h2 id="加锁修改"><a href="#加锁修改" class="headerlink" title="加锁修改"></a><strong>加锁修改</strong></h2><pre class="line-numbers language-java" data-language="java"><code class="language-java">&#x2F;&#x2F;对方法getInstance进行加锁，让其只能同时被一个线程调用
public synchronized Singleton getInstance()
&#x2F;&#x2F;但是这样会导致很大的性能开销，因为加锁只需要在第一次初始化的时候用到其他，都不会再用了。<span aria-hidden="true" class="line-numbers-rows"><span></span><span></span><span></span></span></code></pre>



<h2 id="错误的双重检查锁"><a href="#错误的双重检查锁" class="headerlink" title="错误的双重检查锁"></a>错误的双重检查锁</h2><pre class="line-numbers language-java" data-language="java"><code class="language-java">public class Singleton &#123;
    private static Singleton uniqueSingleton;

    private Singleton() &#123;
    &#125;

    public Singleton getInstance() &#123;
        if (null &#x3D;&#x3D; uniqueSingleton) &#123;
            synchronized (Singleton.class) &#123;
                if (null &#x3D;&#x3D; uniqueSingleton) &#123;
                    uniqueSingleton &#x3D; new Singleton();   &#x2F;&#x2F; error
                &#125;
            &#125;
        &#125;
        return uniqueSingleton;
    &#125;
&#125;<span aria-hidden="true" class="line-numbers-rows"><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span></span></code></pre>

<p>多个线程试图在同一时间创建对象，会通过加锁来保证只有一个线程能创建对象。在对象创建好之后，执行getInstance()方法将不再需要获取锁，直接返回已创建好的对象。</p>
<p>到此看起来很完美，但是这是一个错误优化，<strong>在第一次判断uniqueSingleton不为null的时候可能uniqueSingleton还没有被初始化完成</strong>。</p>
<p><strong>问题的根源</strong></p>
<p>instance = new Singleton(）可以分解为下面三步代码。</p>
<pre class="line-numbers language-java" data-language="java"><code class="language-java">memory &#x3D; allocate();  &#x2F;&#x2F;1.分配对象的内存空间
ctorInstance(memory); &#x2F;&#x2F;2.初始化对象
instance &#x3D; memory;    &#x2F;&#x2F;3.设置instance指向刚刚分配的内存地址<span aria-hidden="true" class="line-numbers-rows"><span></span><span></span><span></span></span></code></pre>

<p>但是编译器可能在在2和3之间重排序。</p>
<p>所以初始化对象代码变成了</p>
<pre class="line-numbers language-java" data-language="java"><code class="language-java">memory &#x3D; allocate();  &#x2F;&#x2F;1.分配对象的内存空间
instance &#x3D; memory;	&#x2F;&#x2F;设置instance指向刚分配的内存地址。但是对象还没有初始化，但是instance已经不为null了。
ctorInstance(memory); &#x2F;&#x2F;初始化对象<span aria-hidden="true" class="line-numbers-rows"><span></span><span></span><span></span></span></code></pre>

<p>为什么这个重排序不会被禁止，因为这三个进行重排序不会改变执行结果，所以这种重排序是可以接受的。</p>
<p><img src="https://gitee.com/modest2001/img/raw/master/yygh/%E5%8F%8C%E9%87%8D%E6%A3%80%E6%9F%A5%E9%94%81%E5%AE%9A2.png" alt="image-20220404153240977"></p>
<h2 id="正确的双重检查锁"><a href="#正确的双重检查锁" class="headerlink" title="正确的双重检查锁"></a>正确的双重检查锁</h2><p>为了解决这种初始流程被重排序，我们使用volatile关键后，重排序被禁止，所有写操作都将发生在读操作之前。</p>
<pre class="line-numbers language-java" data-language="java"><code class="language-java">public class Singleton &#123;
    &#x2F;&#x2F;加入volatile关键字
    private volatile static Singleton uniqueSingleton;

    private Singleton() &#123;
    &#125;

    public Singleton getInstance() &#123;
        if (null &#x3D;&#x3D; uniqueSingleton) &#123;
            synchronized (Singleton.class) &#123;
                if (null &#x3D;&#x3D; uniqueSingleton) &#123;
                    uniqueSingleton &#x3D; new Singleton();
                &#125;
            &#125;
        &#125;
        return uniqueSingleton;
    &#125;
&#125;<span aria-hidden="true" class="line-numbers-rows"><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span></span></code></pre>



<h1 id="基于类初始化的解决方案"><a href="#基于类初始化的解决方案" class="headerlink" title="基于类初始化的解决方案"></a>基于类初始化的解决方案</h1><p>JVM在类的初始化阶段（在Class被加载后，且这个线程被使用之前），会执行类的初始化。在执行类的初始化期间，JVM会去获取一个锁。这个锁可以同步多个线程对同一个类的初始化。</p>
<p>基于这个特性实现了另一种线程安全的延迟初始化方案。（<strong>Initialization On Demand Holder idiom</strong>）</p>
<pre class="line-numbers language-java" data-language="java"><code class="language-java">public class InstanceFactory &#123;
    
    &#x2F;&#x2F;使用静态内部类
    private static class InstanceHolder &#123;
        public static InstanceFactory instance &#x3D; new InstanceFactory();
    &#125;
 
    public static InstanceFactory getInstance() &#123;
        return InstanceHolder.instance;  &#x2F;&#x2F; 这里将导致InstanceHolder类被初始化
    &#125;
&#125;<span aria-hidden="true" class="line-numbers-rows"><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span></span></code></pre>

<p><img src="https://gitee.com/modest2001/img/raw/master/yygh/%E5%9F%BA%E4%BA%8E%E7%B1%BB%E5%88%9D%E5%A7%8B%E5%8C%96%E7%89%B9%E6%80%A7%E7%9A%84%E5%BB%B6%E8%BF%9F%E5%88%9D%E5%A7%8B%E5%8C%96.png" alt="image-20220404161716906"></p>
<p>方案的实质：运行2，3重排序，但是不允许非构造函数线程B“看到”这个重排序。</p>
<p>java语言规范规定：对于每一个类或者接口C，都有一个唯一的初始化锁LC与之对应。从C到LC的映射，由JVM的具体实现去自由实现。JVM在类初始化期间会获取这个初始化锁，并且每一个线程至少获取一次锁来确保这个类已经被初始化过了。</p>
<h2 id="类初始化处理过程中的同步机制（5个阶段）"><a href="#类初始化处理过程中的同步机制（5个阶段）" class="headerlink" title="类初始化处理过程中的同步机制（5个阶段）"></a>类初始化处理过程中的同步机制（5个阶段）</h2><p>核心状态：<code>noinitializion</code> -&gt;  <code>initalizing</code>  -&gt;<code>initalized</code></p>
<p>阶段一：<strong>通过在 <code>Class</code>对象上同步(即获取 <code>Class</code>对象的初始化锁),来控制类或接口的初始化。这个获取锁的线程会一直等待，知道当前线程能够获取到这个初始化锁。</strong></p>
<p><img src="https://gitee.com/modest2001/img/raw/master/yygh/%E7%B1%BB%E5%88%9D%E5%A7%8B%E5%8C%96%E5%90%8C%E6%AD%A5%E7%AC%AC%E4%B8%80%E9%98%B6%E6%AE%B5.png" alt="image-20220404163529382"></p>
<p>A的将类状态从<code>noInitialization</code>变成了<code>initializing</code>。</p>
<p>阶段二：<strong>线程A 执行类的初始化，同时线程B 在初始化锁对应的condition 上等待。</strong></p>
<p><img src="https://gitee.com/modest2001/img/raw/master/yygh/%E7%B1%BB%E5%88%9D%E5%A7%8B%E5%90%8C%E6%AD%A5%E6%9C%BA%E5%88%B62.png" alt="image-20220404163712644"></p>
<p>A的开始初始化，而B读到state=<code>initialzing</code>，然后在初始锁的条件中等待。</p>
<p>阶段三：<strong>线程A 设置 <code>state = initialized</code> ,然后唤醒在 <code>condition</code> 中等待的所有线程。</strong></p>
<p><img src="https://gitee.com/modest2001/img/raw/master/yygh/%E7%B1%BB%E5%88%9D%E5%A7%8B%E5%8C%96%E5%90%8C%E6%AD%A5%E6%9C%BA%E5%88%B63.png" alt="image-20220404163951000"></p>
<p>A已经完成了类的初始化状态从<code>initializing</code>变成了<code>initialized</code>，这时候释放了初始化锁，然后唤醒在condition中的B线程。</p>
<p>阶段四：<strong>线程B 结束类的初始化处理</strong></p>
<p><img src="https://gitee.com/modest2001/img/raw/master/yygh/%E7%B1%BB%E5%88%9D%E5%A7%8B%E5%8C%96%E5%90%8C%E6%AD%A5%E6%9C%BA%E5%88%B64.png" alt="image-20220404164552084"></p>
<p>这时候A已经完成了初始化阶段，但是B读到类的state=<code>initialized</code>表示类已经完成了初始化，所以这时候释放锁，然后完成类初始化阶段完成。（所以B啥也没干，就是等到A初始化完成，然后一直在不停的读取状态）。</p>
<p>阶段五：<strong>线程C执行类的初始化的处理</strong>（和4阶段的B一样，只不过没有B陪跑的久，因为状态就已经变成了initialized）</p>
<p><img src="https://gitee.com/modest2001/img/raw/master/yygh/%E7%B1%BB%E5%88%9D%E5%A7%8B%E5%90%8C%E6%AD%A5%E6%9C%BA%E5%88%B65.png" alt="image-20220404165308659"></p>
<p>C只用了一次获取锁在释放锁。</p>
<p>注意： 这里的 <code>condition</code> 和 <code>state</code> 标记是本文虚构的，Java语言规范并没有硬性规定一定要使用 <code>condition</code> 和 <code>state</code> 标记。<strong>JVM的具体实现只要实现类似的功能即可。</strong></p>
<h1 id="双重检查锁定和类初始化的总结"><a href="#双重检查锁定和类初始化的总结" class="headerlink" title="双重检查锁定和类初始化的总结"></a>双重检查锁定和类初始化的总结</h1><p>总结：两种方式都能够延长初始类，但基于 <code>volatile</code> 的双重检查锁定的方案有一个优势：<strong>除了可以对静态字段实现延迟初始化外，还可以对实例字段实现延迟初始化。</strong> </p>
<p> 字段延迟初始化降低了初始类或创建实例的开销，但增加了访问被延迟化字段的开销。在大多数的时候，正常的初始化要优于延迟初始化。</p>
<p>使用场景：<strong>如果确实对实例字段使用线程安全的延迟初始化，最好使用volatile的双重检查锁定。如果确实对静态字段使用线程安全的延迟初始化，使用类初始化方案。</strong></p>

                
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